What is the degeneracy of the n = 3 level? That is, how many different states are contained in the energy level E3 = -E0/9? Strategy This is a good opportunity to review the quantum numbers we have been discussing. The energy eigenvalues for atomic hydrogen depend only on the principal quantum

Question

What is the degeneracy of the n = 3 level? That is, how many different states are contained in the energy level [latex]E_{3}=-E_{0} / 9[/latex]?

Strategy This is a good opportunity to review the quantum numbers we have been discussing. The energy eigenvalues for atomic hydrogen depend only on the principal quantum number n (in the absence of a magnetic field). For each value of n, there can be n different orbital angular momentum [latex]\ell ext { states }(\ell=0,1, \ldots, n-1)[/latex]. For each value of [latex]\ell ext {, there are } 2 \ell+1[/latex] different magnetic quantum states [latex]\left(m_{\ell}=-\ell,-\ell+1, \ldots, 0,1, \ldots,+\ell ight)[/latex].

Accepted Answer

To find the total degeneracy for n = 3 we have to add all the possibilities.

n	[latex]\ell[/latex]	[latex]m_{\ell}[/latex]	[latex]2 \ell+1[/latex]
3	0	0	1
3	1	[latex]-1,0,1[/latex]	3
3	2	[latex]-2,-1,0,1,2[/latex]	5
			Total = 9

The n = 3 level is degenerate (in the absence of a magnetic field) because all nine states have the same energy but different quantum numbers. Their wave functions, however, are quite different. You may notice that, in general, the degeneracy is [latex]n^{2}[/latex] (see Problem 16).

Question 7.4: What is the degeneracy of the n = 3 level? That is, how many...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions